%minor changes made by KC 11/10/99 \documentclass[12pt]{amsart} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amscd} %TCIDATA{TCIstyle=article/art1.lat,amsart,amsart} %TCIDATA{Created=Fri Oct 11 19:19:45 1996} %TCIDATA{LastRevised=Sat Dec 05 19:06:41 1998} %TCIDATA{Language=American English} \textwidth 6in \oddsidemargin0.25in \evensidemargin0.25in \renewcommand{\baselinestretch}{1.4} \newcommand{\real}{\Bbb{R}} \newcommand{\AC}{{\rm AC}} \newcommand{\Conn}{\mathcal{C}} \newcommand{\darb}{\mathcal{D}} \newcommand{\Ext}{{\rm Ext}} \newcommand{\PC}{{\rm PC}} \newcommand{\bd}{{\rm bd}} \newcommand{\e}{\varepsilon} \newtheorem{problem}{Problem} \input tcilatex \begin{document} \title{Topological Dimension and Sums of Connectivity Functions} \author{Krzysztof Ciesielski and Jerzy Wojciechowski} \address{Department of Mathematics\\ West Virginia University\\ PO Box 6310\\ Morgantown, WV 26506-6310, USA} \email[K. Ciesielski]{K\_Cies@math.wvu.edu \endgraf {\it Web page address,} K. Ciesielski: {\tt http://www.math.wvu.edu/homepages/kcies}} \email[J. Wojciechowski]{jerzy@math.wvu.edu} \subjclass{Primary 54F45; Secondary 54C30, 26B40} \keywords{inductive dimension, connectivity functions, Darboux functions} \thanks{This work was partially supported by NSF Cooperative Research Grant INT-9600548 with its Polish part being financed by Polish Academy of Science PAN} \maketitle \begin{abstract} The main goal of this paper is to show that the inductive dimension of a $% \sigma $-compact metric space $X$ can be characterized in terms of algebraical sums of connectivity (or Darboux) functions $X\rightarrow \Bbb{R} $. As an intermediate step we show, using a result of Hayashi~\cite{Hayashi: axiomatic dimension euclidean}, that for any dense $G_{\delta }$ set $G\in \Bbb{R}^{2k+1}$ the union of $G$ and some $k$ homeomorphic images of $G$ is universal for $k$-dimensional separable metric spaces. We will also discuss how our definition works with respect to other classes of Darboux-like functions. In particular, we show that for the class of peripherally continuous functions on an arbitrary separable metric space $X$ our parameter is equal to either $\limfunc{ind}X$ or $\limfunc{ind}X-1$. Whether the later is at all possible, is an open probem. \end{abstract} \section{Introduction} Our terminology and notation is standard and follows~\cite{Ciesielski: set theory for working}. Let $X$ be a non-empty set and $\mathcal{F}$ be a family of functions from $X$ into $\Bbb{R}$. If $m$ is a nonnegative integer, then let \begin{equation*} m\mathcal{F}=\left\{ f_{1}+\dots +f_{m}:f_{1},\dots ,f_{m}\in \mathcal{F}% \right\} , \end{equation*} and let $\Bbb{R}^{X}$ be the family consisting of all functions from $X$ into $\Bbb{R}$. Let $\limfunc{DIM}\nolimits_{\mathcal{F}}X$ be defined by \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{F}}X=\min \left( \left\{ m\in \Bbb{Z}:m\ge 0% \text{ and }\left( m+1\right) \mathcal{F}=\Bbb{R}^{X}\right\} \cup \left\{ \infty \right\} \right) . \end{equation*} Given a metric space $X$, a function $f:X\rightarrow \Bbb{R}$ is a \emph{% connectivity function} (\emph{Darboux function}) if for every connected subset $C$ of $X$ the graph of the restriction $f|C$ is a connected subset of $X\times \Bbb{R}$ (the image $f[C]$ is connected in $\Bbb{R}$). The following theorem holds. \begin{theorem} \label{thm: connectivity on Rn}If $n$ is a positive integer and $\mathcal{F}$ is the family of connectivity functions or the family of Darboux functions on $\Bbb{R}^{n}$, then \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{F}}\Bbb{R}^{n}=n. \end{equation*} \end{theorem} The proof of Theorem \ref{thm: connectivity on Rn} is given by Ciesielski and Wojciechowski \cite{CieWoj: sums of connectivity}, except for the case $% n=1$ that has been proved by Ciesielski and Rec\l aw \cite{CieRec: invariants concerning extendable and}, and the inequality $\ge $ in the case of Darboux functions that has been demonstrated by Jordan \cite{Jordan: PhD,Jordan:darb}. Theorem \ref{thm: connectivity on Rn} motivates the notation $\limfunc{DIM}% \nolimits_{\mathcal{F}}X$ and shows that (with suitably chosen family $% \mathcal{F}$) $\limfunc{DIM}\nolimits_{\mathcal{F}}X$ can be considered as a sort of dimension of $X$ (\emph{dimension relative to} $\mathcal{F}$). In this paper we are going to show that the dimension relative to the family of connectivity (Darboux) functions coincides with the inductive dimension $% \func{ind}$ on every $\sigma $-compact metric space. Let $X$ be a separable metric space. Given $A,B\subseteq X$, the \emph{% boundary} of $A\cap B$ in $A$ will be denoted by $\limfunc{bd}_{A}B$. The \emph{inductive dimension} $\func{ind}A$ of a subset $A\subseteq X$ is defined inductively as follows. (See for example Engelking \cite{Engelking: general topology}.) \begin{enumerate} \item[(i)] $\func{ind}A=-1$ if and only if $A=\emptyset $. \item[(ii)] $\func{ind}A\le m$ if for any $p\in A$ and any open neighborhood $W$ of $p$ there exists an open neighborhood $U\subseteq W$ of $% p$ such that $\func{ind}\limfunc{bd}_{A}U\le m-1$. \item[(iii)] $\func{ind}A=m$ if $\func{ind}A\le m$ and it is not true that $% \func{ind}A\le m-1$. \end{enumerate} \noindent Let $\mathcal{C}$ be the family of connectivity functions on $X$ and $\mathcal{D}$ be the family of Darboux functions on $X$. Our main result is the following theorem. \begin{theorem} \label{thm: dimension}If $X$ is a $\sigma $-compact metric space, then \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=% \func{ind}X. \end{equation*} \end{theorem} Clearly \begin{equation} \limfunc{DIM}\nolimits_{\mathcal{F}}X\ge \limfunc{DIM}\nolimits_{\mathcal{G}% }X\quad \text{for any }\mathcal{F}\subseteq \mathcal{G}\subseteq \Bbb{R}^{X} \label{eq1} \end{equation} Since $\mathcal{C}\subseteq \mathcal{D}$ for any space $X$, we have $% \limfunc{DIM}\nolimits_{\mathcal{C}}X\ge \limfunc{DIM}\nolimits_{\mathcal{D}% }X$, and so Theorem \ref{thm: dimension} follows immediately from the following two results. \begin{theorem} \label{thm: upper bound}If $X$ is a separable metric space, then \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{C}}X\le \func{ind}X. \end{equation*} \end{theorem} \begin{theorem} \label{thm: lower bound}If $X$ is a $\sigma $-compact metric space, then \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{D}}X\ge \func{ind}X. \end{equation*} \end{theorem} A natural question is whether Theorem \ref{thm: lower bound} can be extended to all separable metric spaces or perhaps all that are complete. The answer is `no' in both cases since Mazurkiewicz \cite{Mazurkiewicz: problemes Urysohn} has shown that for each positive integer $n$ there exists a complete separable metric space $X$ of inductive dimension $n$ which is totally disconnected, that is, single points are its only connected subspaces. (See also \cite[Example~II~16]{HurWall: dimension theory}.) Since for every totally disconnected space $X$ we have \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=0 \end{equation*} (any function $f:X\rightarrow \Bbb{R}$ is a connectivity and Darboux), we get \begin{equation} \label{eq2} \limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X =00$. It might be interesting to answer the question whether the equation \begin{equation} \label{eq3} \limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X \end{equation} holds for all separable metric spaces $X$ or at least all that are complete. To prove Theorem \ref{thm: upper bound} we will prove the following result which seems to be of independent interest. We say that a separable metric space $X$ is $m$\emph{-dimensional} if $\func{ind}X=m$. If $Y$ is a metric space such that for every $m$-dimensional separable metric space $X$ there is a subspace of $Y$ homeomorphic to $X$, then we say that $Y$ is \emph{% universal\/} for $m$-dimensional separable metric spaces. \begin{theorem} \label{thm: union is universal}If $G$ is a dense $G_{\delta }$ set in $\Bbb{R% }^{2k+1}$, then there are homeomorphisms $h_{j}:\Bbb{R}^{2k+1}\rightarrow \Bbb{R}^{2k+1}$, for $j=1,\dots ,k$, such that $G\cup \bigcup_{j=1}^{k}h_{j}[G]$ is universal for $k$-dimensional separable metric spaces. \end{theorem} Theorem~\ref{thm: union is universal} will be used to prove the following fact, that easily implies Theorem \ref{thm: upper bound}. \begin{proposition} \label{prop} For every positive integer $k$ there exists a dense $G_{\delta } $-set $H$ in $\Bbb{R}^{2k+1}$ such that \begin{itemize} \item[(i)] $H$ is universal for $k$-dimensional separable metric spaces, and \item[(ii)] for every $\varphi \colon\Bbb{R}^{2k+1}\to \Bbb{R}$ there are connectivity functions $g_{0},\ldots ,g_{k}\colon\Bbb{R}^{2k+1}\to \Bbb{R}$ such that $(g_{0}+\cdots +g_{k})(x)=\varphi (x)$ for every $x\in H$. \end{itemize} \end{proposition} The proof of Theorem \ref{thm: union is universal} will be based on Lemma~% \ref{lem: implicit} and Theorem~\ref{thm: negligible G-delta}, that are proved in \cite{CieWoj: sums of connectivity}, and on Theorem \ref{thm: hayashi}, which is proved by Hayashi~\cite{Hayashi: axiomatic dimension euclidean}. Theorem \ref{thm: union is universal} is proved in Section \ref {sec: universal}, the proof of Theorem \ref{thm: upper bound} is presented in Section \ref{sec: upper}, while Theorem \ref{thm: lower bound} is proved in Section \ref{sec: lower}. The authors would like to thank Roman Pol for directing their attention to the results of Hayashi~\cite{Hayashi: axiomatic dimension euclidean} and Mazurkiewicz \cite{Mazurkiewicz: problemes Urysohn}. \section{\label{sec: universal}A $k$-dimensional universal set} In this section we are going to present a proof of Theorem \ref{thm: union is universal}. Let a \emph{countable dense grid in }$\Bbb{R}^{n}$ be a product $B_{1}\times \dots \times B_{n}\subseteq \Bbb{R}^{n}$ where $B_{1},\dots ,B_{n}$ are countable dense subsets of $\Bbb{R}$. If $B=B_{1}\times \dots \times B_{n}$ is a countable dense grid in $\Bbb{R}^{n}$ and $i\le n$, then let $B^{\left( i\right) }$ consist of those points in $\Bbb{R}^{n}$ that differ from a point in $B$ at at most $i$ coordinates, that is, \begin{equation*} B^{\left( i\right) }=\left\{ \left\langle x_{1},\dots ,x_{n}\right\rangle \in \Bbb{R}^{n}:\left| \left\{ j:x_{j}\notin B_{j}\right\} \right| \le i\right\} . \end{equation*} Note that in particular $B^{\left( 0\right) }=B$. Let $\Bbb{Q}$ be the set of rational numbers and $I$ be the closed interval $\left[ 0,1\right] $. Our proof of Theorem \ref{thm: union is universal} uses the following result of Hayashi \cite{Hayashi: axiomatic dimension euclidean}. (See also \cite{GS} for similar results.) \begin{theorem} \label{thm: hayashi}If $G$ is a $G_{\delta }$-set in $I^{2k+1}$ containing $% \left( \Bbb{Q}^{2k+1}\right) ^{\left( k\right) }\cap I^{2k+1}$, then $G$ is universal for $k$-dimensional separable metric spaces. \end{theorem} First notice that Theorem \ref{thm: hayashi} implies immediately the following corollary. \begin{corollary} \label{cor: of hayashi}If $B$ is a countable dense grid in $\Bbb{R}^{2k+1}$ and $G$ is a $G_{\delta }$-set in $\Bbb{R}^{2k+1}$ containing $B^{\left( k\right) }$, then $G$ is universal for $k$-dimensional separable metric spaces. \end{corollary} \begin{proof} Let $B=B_{1}\times \dots \times B_{2k+1}$. Let $g_{1},\dots ,g_{2k+1}:\Bbb{R}% \rightarrow \Bbb{R}$ be increasing homeomorphisms such that $B_{i}=g_{i}[% \Bbb{Q}]$ and \begin{equation*} g=g_{1}\times \dots \times g_{2k+1}:\Bbb{R}^{2k+1}\rightarrow \Bbb{R}^{2k+1}. \end{equation*} Then \begin{equation*} \left( \Bbb{Q}^{2k+1}\right) ^{\left( k\right) }\cap I^{2k+1}\subseteq g^{-1}[G]\cap I^{2k+1}. \end{equation*} Let $X$ be a $k$-dimensional separable metric space. It follows from Theorem \ref{thm: hayashi} that there is a subspace $Y$ of $g^{-1}[G]\cap I^{2k+1}$ that is homeomorphic to $X$. Then $g[Y]$ is a subspace of $G$ that is homeomorphic to $X$. \end{proof} To prove Theorem \ref{thm: union is universal} we will also need a result proved implicitly in \cite{CieWoj: sums of connectivity}. We will first introduce the notation used there. If $\langle B_{i}:i\in n\rangle $ is a family of subsets of $\Bbb{R}$ and $f$ is a function from $\left\{ 1,\dots ,n\right\} $ into $\left\{ 0,1\right\} $, then let \begin{equation*} \prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{R}}\right) =B_{1}^{\prime }\times \dots \times B_{n}^{\prime }, \end{equation*} where \begin{equation*} B_{i}^{\prime }=\left\{ \begin{array}{cl} B_{i} & \text{if }f(i)=0, \\ \Bbb{R} & \text{if }f(i)=1. \end{array} \right. \end{equation*} The following lemma is stated implicitly and proved in \cite{CieWoj: sums of connectivity} (the inductive condition (8) in the proof of Proposition~2.4, page 419). \begin{lemma} \label{lem: implicit}If $G$ is a dense $G_{\delta }$-set in $\Bbb{R}^{n}$, then there are countable dense sets $B_{i}\subseteq \Bbb{R}$ and homeomorphisms $h_{i}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$, for $i=1,\dots ,n$% , such that \begin{equation*} \prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{R}}\right) \subseteq G\cup \bigcup_{i=1}^{k}h_{i}[G] \end{equation*} for every $k\in \left\{ 0,1,\dots ,n\right\} $ and every function $f:\left\{ 1,\dots ,n\right\} \to \left\{ 0,1\right\} $ such that $\left| f^{-1}(1)\right| =k$. \end{lemma} Lemma \ref{lem: implicit} implies immediately the following result. \begin{theorem} \label{thm: cover grid}If $G$ is a dense $G_{\delta }$-set in $\Bbb{R}^{n}$ and $k\le n$, then there is a countable dense grid $B$ in $\Bbb{R}^{n}$ and homeomorphisms $h_{1},\dots ,h_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ such that \begin{equation*} B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G]. \end{equation*} \end{theorem} \begin{proof} Let $G$ be a dense $G_{\delta }$-set in $\Bbb{R}^{n}$. For $i=1,\dots ,n$, let $B_{i}\subseteq \Bbb{R}$ be countable dense sets and $h_{i}:\Bbb{R}% ^{n}\rightarrow \Bbb{R}^{n}$ be homeomorphisms as in Lemma \ref{lem: implicit}. Then \begin{equation*} B=B_{1}\times \dots \times B_{n} \end{equation*} is a countable dense grid in $\Bbb{R}^{n}$ and \begin{equation*} B^{(k)}=\bigcup \left\{ \prod_{i=1}^{n}\left( B_{i}\vee _{f}\QTR{userA}{\Bbb{% R}}\right) :\left| f^{-1}(1)\right| =k\right\} . \end{equation*} It follows from Lemma \ref{lem: implicit} that \begin{equation*} B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G]. \end{equation*} \end{proof} \noindent \textbf{Proof of Theorem \ref{thm: union is universal}. }Let $G$ be a dense $G_{\delta}$-set in $\Bbb{R}^{2k+1}$. By Theorem \ref{thm: cover grid}, there is a countable dense grid $B$ in $\Bbb{R}^{2k+1}$ and homeomorphisms $h_{1},\dots ,h_{k}:\Bbb{R}^{2k+1}\rightarrow \Bbb{R}^{2k+1}$ such that $B^{\left( k\right) }\subseteq G\cup \bigcup_{j=1}^{k}h_{j}[G]$. By Corollary \ref{cor: of hayashi}, $G\cup \bigcup_{j=1}^{k}h_{j}[G]$ is universal for $k$-dimensional separable metric spaces. $\blacksquare $\medskip \section{\label{sec: upper}Inductive dimension as the upper bound} Now we shall prove Theorem \ref{thm: upper bound}. Beside Theorem \ref{thm: union is universal} we will need the following result. (See \cite[Proposition 2.3]{CieWoj: sums of connectivity}). \begin{theorem} \label{thm: negligible G-delta}For every $n>1$, there exists a function $f:% \QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ and a dense $% G_{\delta }$ subset $G$ of $\QTR{userA}{\Bbb{R}}^{n}$ such that any function $g:\QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ with $g(x)=f(x)$ for $x\notin G$ is a connectivity function. \end{theorem} Let us now introduce some notation. If $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$ and $A\subseteq \Bbb{R}^{n}$, then we will write $g\equiv _{A}f$ if and only if $g(x)=f(x)$ for every $x\in \Bbb{R}^{n}\setminus A$. Notice that if $% g\equiv _{A}f$ and $A\subseteq A^{\prime }$, then $g\equiv _{A^{\prime }}f$. Also $g\equiv _{\emptyset }f$ if and only if $g=f$, and $g\equiv _{\Bbb{R}% ^{n}}f$ for any $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$. The following two lemmas are easy observations. \begin{lemma} \label{lem: simple property1}Let $f,g:\Bbb{R}^{n}\rightarrow \Bbb{R}$ and $% A\subseteq \Bbb{R}^{n}$. If $h:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ is a bijection, then $g\equiv _{h[A]}\left( f\circ h^{-1}\right) $ if and only if $\left( g\circ h\right) \equiv _{A}f$. \end{lemma} \begin{proof} Assume $g\equiv _{h[A]}\left( f\circ h^{-1}\right) $. Then $% g(x)=f(h^{-1}(x)) $ for every $x\in \Bbb{R}^{n}\setminus h[A]$. If $y\in \Bbb{R}^{n}\setminus A $, then $h(y)\in \Bbb{R}^{n}\setminus h[A]$ so \begin{equation*} \left( g\circ h\right) (y)=g(h(y))=f(h^{-1}(h(y)))=f(y), \end{equation*} implying that $\left( g\circ h\right) \equiv _{A}f$. The opposite implication is proved similarly. \end{proof} \begin{lemma} \label{lem: simple property2}Let $g_{0}^{\prime },\dots ,g_{k}^{\prime }:% \Bbb{R}^{n}\rightarrow \Bbb{R}$. If $A\subseteq \Bbb{R}^{n}$, and $\left\{ A_{0},\dots ,A_{k}\right\} $ is a partition of $A$, then for any $\varphi :A\rightarrow \Bbb{R}$ there are $g_{0},\dots ,g_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}$ such that \begin{equation*} g_{i}\equiv _{A_{i}}g_{i}^{\prime },\quad \quad i=0,\dots ,k, \end{equation*} and the restriction of $g_{0}+\dots +g_{k}$ to $A$ is equal to $\varphi $. \end{lemma} \begin{proof} Define $g_{i}:\Bbb{R}^{n}\rightarrow \Bbb{R}$ by \begin{equation*} g_{i}(x)=\left\{ \begin{array}{ll} \varphi (x)-\sum_{j\neq i}g_{j}^{\prime }(x) & \text{if }x\in A_{i}, \\ g_{i}^{\prime }(x) & \text{if }x\notin A_{i}. \end{array} \right. \end{equation*} Then $\varphi (x)=g_{0}(x)+\dots +g_{k}(x)$ for every $x\in A$. \end{proof} \noindent \textbf{Proof of Proposition \ref{prop}.} Let $n=2k+1$. By Theorem \ref{thm: negligible G-delta} there exists a function $f\colon \QTR{userA}{% \Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ and a dense $G_{\delta }$% -subset $G$ of $\QTR{userA}{\Bbb{R}}^{n}$ such that any function $g\colon \QTR{userA}{\Bbb{R}}^{n}\rightarrow \QTR{userA}{\Bbb{R}}$ with $g\equiv_{G}f$ is a connectivity function. By Theorem \ref{thm: union is universal}, there are homeomorphisms $h_{i}\colon \Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$, for $% i=1,\ldots ,k$, such that the $G_\delta$ set $H=G\cup \bigcup_{j=1}^{k}h_{j}[G]$ is universal for $k$-dimensional separable metric spaces. Let $\left\{ A_{0},\ldots ,A_{k}\right\} $ be the partition of $H$ defined inductively by \begin{equation*} A_{0}=G,\quad A_{j}=h_{j}[G]\setminus \left( A_{0}\cup \cdots \cup A_{j-1}\right) ,\quad j=1,\dots ,k. \end{equation*} Let $\varphi\colon\Bbb{R}^{n}\rightarrow \Bbb{R}$ be an arbitrary function, and $h_{0}:\Bbb{R}^{n}\rightarrow \Bbb{R}^{n}$ be the identity function. It follows from Lemma \ref{lem: simple property2}, that there are functions $% g_{0},\dots ,g_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R}$ such that \begin{equation*} g_{i}\equiv _{A_{i}}\left( f\circ h_{i}^{-1}\right) ,\quad \quad i=0,\dots ,k, \end{equation*} and the restriction of $g_{0}+\dots +g_{k}$ to $H$ is equal to $% \varphi\restriction H$. It remains to prove that $g_i$'s are connectivity functions. Let $i\in \left\{ 0,\dots ,k\right\} $. Since $A_{i}\subseteq h_{i}[G]$, we have \begin{equation*} g_{i}\equiv _{h_{i}[G]}\left( f\circ h_{i}^{-1}\right) , \end{equation*} and so Lemma \ref{lem: simple property1} implies that \begin{equation*} g_{i}\circ h_{i}\equiv _{G}f. \end{equation*} Thus $g_{i}\circ h_{i}$ (and hence $g_{i}$) is a connectivity function on $% \Bbb{R}^{n}$. $\blacksquare $ \medskip \noindent \textbf{Proof of Theorem \ref{thm: upper bound}.} Let $X$ be a $k$% -dimensional separable metric space. If $k=0$, then any function $% X\rightarrow \Bbb{R}$ is a connectivity function, so we can assume that $% k\ge 1$. Let $H$ be a $G_{\delta }$-set from Proposition~\ref{prop}. Then there is a subspace $A$ of $H$ homeomorphic to $X$. Take an arbitrary $% \varphi _{0}\colon A\to \Bbb{R}$. We have to show that $\varphi _{0}$ is a sum of $k+1$ connectivity functions on $A$. Let $\varphi \colon\Bbb{R}^{n}\rightarrow \Bbb{R}$ be an arbitrary extension of $\varphi _{0}$ and let $g_{0},\ldots ,g_{k}:\Bbb{R}^{n}\rightarrow \Bbb{R} $ be connectivity functions such that $(g_{0}+\cdots +g_{k})(x)=\varphi (x)$ for all $x\in H$. Then the functions $g_{i}\upharpoonright A$ are connectivity and $(g_{0}\upharpoonright A)+\cdots +(g_{k}\upharpoonright A)=\varphi _{0}$. $\blacksquare $ \section{\label{sec: lower}Inductive dimension as the lower bound} In this section we are going to prove Theorem \ref{thm: lower bound}. In the proof that follows we will need some additional definitions and results from dimension theory. (See for example~\cite{HurWall: dimension theory}). \begin{lemma} \label{lem: countable union}If $X$ is a separable metric space and \begin{equation*} X=\bigcup_{i=1}^{\infty }X_{i}, \end{equation*} where $X_{i}$ is closed in $X$ and $\limfunc{ind}X_{i}\le m$, for $% i=1,2,\dots $, then $\limfunc{ind}X\le m$. \end{lemma} Given $X\subseteq \QTR{userA}{\Bbb{R}}^{n}$ and an integer $m\ge 1$, we say that $X$ is an \emph{$m$-dimensional Cantor-manifold} if $X$ is compact, $% \limfunc{ind}X=m$, and for every $Y\subseteq X$ with $\limfunc{ind}Y\le m-2$% , the set $X\setminus Y$ is connected. The following lemma is proved in~\cite{HurWall: dimension theory}. \begin{lemma} \label{lem: Cantor inside}For any compact $Y\subseteq \QTR{userA}{\Bbb{R}}% ^{n}$ with $\limfunc{ind}Y\ge m$ there exists an $m$-dimensional Cantor manifold $X\subseteq Y$. \end{lemma} We will also need the following result of Francis Jordan. (See \cite[Lemma 3.3.8]{Jordan: PhD} or \cite[Lemma 3.8]{Jordan:darb}.) A \emph{% perfect set} is a non-empty closed set without isolated points. \begin{lemma} \label{lem: Jordan}Let $n>1$ and $M$ be an $n$-dimensional Cantor manifold. If $n\ge k\ge 1$ and $f\in k\mathcal{D}$, where $\mathcal{D}$ is the family of Darboux functions $M\rightarrow \Bbb{R}$, then there is a connected perfect set $P\subseteq M$ such that the restriction of $f$ to $P$ is Darboux. \end{lemma} A \emph{Bernstein set}, is a set $B\subseteq \QTR{userA}{\Bbb{R}}^{n}$ such that $B\cap P\neq \emptyset $ and $B\setminus P\neq \emptyset $ for every perfect set $P\subseteq \QTR{userA}{\Bbb{R}}^{n}$. Note that the characteristic function of a Bernstein set is not Darboux on any perfect set. Now we are ready to prove Theorem \ref{thm: lower bound}.\medskip \textbf{Proof of Theorem~\ref{thm: lower bound}. }Suppose, by way of contradiction, that there exists a $k$-dimensional $\sigma $-compact metric space $X$ such that \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{D}}X<\func{ind}X=k, \end{equation*} where $\mathcal{D}$ is the family of Darboux functions on $X$. We can assume that $X\subseteq \Bbb{R}^{m}$ for some positive integer $m$. Then \begin{equation*} X=\bigcup_{i=1}^{\infty }X_{i}, \end{equation*} with $X_{i}$ compact, $i=1,2\dots $ and it follows from Lemma \ref{lem: countable union} that there is a positive integer $j$ with $\limfunc{ind}% X_{j}\ge k$. By Lemma \ref{lem: Cantor inside} there is a $k$-dimensional Cantor manifold $M\subseteq X_{j}$. Let $B\in \Bbb{R}^{m}$ be a Bernstein set and $f:X\rightarrow \Bbb{R}$ be the characteristic function of $B\cap X$. Since $\limfunc{DIM}\nolimits_{% \mathcal{D}}X1$, %TCIMACRO{ %\TeXButton{inclusions for n>1}{\begin{picture}(0,90) % \put(130,55){\makebox(0,0){$\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)$}} % \put(205,55){\vector(1,0){15}} % \put(270,55){\makebox(0,0){$\AC(\real^n)\cap\darb(\real^n)$}} % \put(320,60){\vector(2,1){18}} % \put(362,70){\makebox(0,0){${\AC(\real^n)}$}} % \put(360,40){\makebox(0,0){${\darb(\real^n)}$}} % \put(320,52){\vector(2,-1){18}} %\end{picture} %\vskip-18pt %}} %BeginExpansion \begin{picture}(0,90) \put(130,55){\makebox(0,0){$\Ext(\real^n)=\Conn(\real^n)=\PC(\real^n)$}} \put(205,55){\vector(1,0){15}} \put(270,55){\makebox(0,0){$\AC(\real^n)\cap\darb(\real^n)$}} \put(320,60){\vector(2,1){18}} \put(362,70){\makebox(0,0){${\AC(\real^n)}$}} \put(360,40){\makebox(0,0){${\darb(\real^n)}$}} \put(320,52){\vector(2,-1){18}} \end{picture} \vskip-18pt % %EndExpansion \noindent where $\vvv$ denotes a strict inclusion. Natkaniec~\cite[prop. 1.7.1]{AC} proved that every function $f\colon \Bbb{R}% ^{n}\to \Bbb{R}$ is a sum of two almost continuous functions. This implies that \begin{equation} \limfunc{DIM}\nolimits_{\limfunc{AC}}\Bbb{R}^{n}=1\text{\quad for every }% n=1,2,3,\ldots \end{equation} making the class $\limfunc{AC}$ useless in our definition of dimension. The situation is different for the remaining two classes. Let $X$ be a separable metric space. Since \begin{equation*} \limfunc{Ext}(\Bbb{R}^{2k+1})=\mathcal{C}(\Bbb{R}^{2k+1})=\limfunc{PC}(\Bbb{R% }^{2k+1}) \end{equation*} for $k\geq 1$, and since any function $X\rightarrow \Bbb{R}$ is both peripherally continuous and extendable when $\limfunc{ind}X=0$, the inequalities \begin{equation} \limfunc{DIM}\nolimits_{\limfunc{Ext}}X\leq \func{ind}X\quad \text{and}\quad \limfunc{DIM}\nolimits_{\limfunc{PC}}X\leq \func{ind}X \label{eq11} \end{equation} follow from Proposition~\ref{prop} in precisely the same way as Theorem~\ref {thm: upper bound} does. Moreover, it is immediate to see that the analog of Theorem~\ref{thm: dimension} for the class $\limfunc{Ext}$ is also true. \begin{theorem} \label{thm:dimension2} If $X$ is a $\sigma $-compact metric space, then \begin{equation*} \limfunc{DIM}\nolimits_{\limfunc{Ext}}X=\func{ind}X. \end{equation*} \end{theorem} \begin{proof} The inequality $\limfunc{DIM}\nolimits_{\limfunc{Ext}}X\leq \func{ind}X$ is a restatement of (\ref{eq11}). The other inequality holds since for every $% \sigma $-compact metric space~$X$ we have $\limfunc{DIM}\nolimits_{\mathcal{C% }}X=\func{ind}X$ and the inequality $\limfunc{DIM}\nolimits_{\limfunc{Ext}% }X\ge \limfunc{DIM}\nolimits_{\mathcal{C}}X$ is implied by $\limfunc{Ext}% (X)\subseteq \mathcal{C}(X)$ and (\ref{eq1}). \end{proof} In the case of the class $\limfunc{PC}$ the situation is quite different. Unlike for the classes $\mathcal{C}$, $\mathcal{D}$, and $\limfunc{Ext}$, (see (\ref{eq2}) which holds also for $\limfunc{DIM}\nolimits_{\limfunc{Ext}% }X$) the dimension relative to the class $\limfunc{PC}$ is very close to the inductive dimension for every separable metric space. However, it is not clear whether we have equality even for all compact metric spaces. %On one side, there are no Polish spaces that %make the number %$\limfunc{DIM}\nolimits_{\PC}X$ %as useless as the example from (\ref{eq2}) %makes in case of numbers %$\limfunc{DIM}\nolimits_{\mathcal{C}}X$ and %$\limfunc{DIM}\nolimits_{\mathcal{D}}X$. %On the other side, we do not know %whether the equation %$\limfunc{DIM}\nolimits_{\PC}X=\func{ind}X$ %holds, in general, even for the compact metric spaces. \begin{theorem} \label{thm:seq5} If $X$ is a separable metric space, then \begin{equation} \limfunc{ind}X-1\leq \limfunc{DIM}\nolimits_{\limfunc{PC}}X\leq \limfunc{ind}% X. \label{11} \end{equation} \end{theorem} \begin{proof} Let $k=\limfunc{ind}X$. The inequality $\limfunc{DIM}\nolimits_{\limfunc{PC}% }X\leq k$ is a restatement of (\ref{eq11}). To prove the other inequality we will show that \begin{itemize} \item[($*$)] for every $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$ and $% \varepsilon >0$ there exist a closed subset $Y$ of $X$ of cardinality continuum such that \begin{equation*} \left| g_{i}(x)-g_{i}(y)\right| <\varepsilon \end{equation*} for every $x,y\in Y$ and $i=1,2,\dots ,k-1$. \end{itemize} We prove ($*$) by induction on $k\ge 1$. If $k=1$, take $Y=X$. The cardinality of $X$ cannot be smaller than continuum since for some $x\in X$ and $r>0$ the boundaries of the open balls in $X$ with center $x$ and radius smaller than $r$ are non-empty and pairwise disjoint. Assume that $k\ge 2$. Let $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$ and $% \varepsilon >0$. There is $p\in X$ and an open neighborhood $W$ of $p$ such that $\limfunc{ind}\limfunc{bd}(U)=k-1$ for any open $U$ with $p\in U\subseteq W$. Since $g_{1}$ is peripherally continuous, there is an open neighborhood $U$ of $p$ such that $U\subseteq W$ and the image $g_{1}[% \limfunc{bd}(U)]$ is contained in the open interval $\left( g_{1}(p)-\varepsilon /2,g_{1}(p)+\varepsilon /2\right) $. Since $\limfunc{ind% }\limfunc{bd}(U)=k-1$, it follows from the inductive hypothesis that there is a closed subset $Y$ of $\limfunc{bd}(U)$ of cardinality continuum such that \begin{equation} \left| g_{i}(x)-g_{i}(y)\right| <\varepsilon \label{abc} \end{equation} for every $x,y\in Y$ and $i=2,3,\dots ,k-1$. Then $Y$ is closed in $X$ and it follows from the choice of $U$, that (\ref{abc}) holds also for $i=1$ completing the proof of ($*$). Now we show that ($*$) implies that \begin{equation*} \limfunc{DIM}\nolimits_{\limfunc{PC}}X\ge k-1. \end{equation*} Let $Z$ be a subset of $X$ such that $A\cap Z\neq \emptyset $ and $% A\setminus Z\neq \emptyset $ for every closed $A\subseteq X$ of cardinality continuum. The existence of such $Z$ can be proved by listing all closed subsets of $X$ of cardinality continuum in a sequence $\langle A_{\alpha }\rangle _{\alpha <% \frak{c}}$ of length continuum, defining two sequences $\langle a_{\alpha }\rangle _{\alpha <\mathfrak{c}}$ and $\langle b_{\alpha }\rangle _{\alpha <\frak{% c}}$ of points in $X$ by transfinite induction so that \begin{equation*} a_{\alpha }\in A_{\alpha }\setminus \left( \left\{ a_{\beta }:\beta <\alpha \right\} \cup \left\{ b_{\beta }:\beta <\alpha \right\} \right) , \end{equation*} and \begin{equation*} b_{\alpha }\in A_{\alpha }\setminus \left( \left\{ a_{\beta }:\beta \le \alpha \right\} \cup \left\{ b_{\beta }:\beta <\alpha \right\} \right) , \end{equation*} for every $\alpha <\mathfrak{c}$, and putting \begin{equation*} Z=\left\{ a_{\alpha }:\alpha <\frak{c}\right\} . \end{equation*} Let $f:X\rightarrow \Bbb{R}$ be the characteristic function of the set $Z$. The proof will be complete when we show that \begin{equation*} f\notin \left( k-1\right) \limfunc{PC}(X). \end{equation*} Suppose, by way of contradiction, that \begin{equation*} f=g_{1}+\cdots +g_{k-1} \end{equation*} for some $g_{1},\ldots ,g_{k-1}\in \limfunc{PC}(X)$. By ($*$) there is a closed subset $Y$ of $X$ of cardinality continuum such that \begin{equation*} \left| g_{i}(x)-g_{i}(y)\right| <\frac{1}{k-1} \end{equation*} for every $x,y\in Y$ and $i=1,2,\dots ,k-1$. Therefore \begin{equation*} \left| f(x)-f(y)\right| <1 \end{equation*} for every $x,y\in Y$. Since $Y\cap Z$ and $Y\setminus Z$ are both non-empty, there are $x,y\in Y$ with $f(x)=0$ and $f(y)=1$ and we get a contradiction. Thus the proof is complete. \end{proof} \begin{corollary} If $X$ is a space of Mazurkiewicz of dimension $k\ge 2$, then the class $% \limfunc{PC}(X)$ is not equal to either $\mathcal{C}(X)$, $\mathcal{D}(X)$, or $\limfunc{Ext}(X)$. \end{corollary} \begin{proof} If $X$ is a space of Mazurkiewicz of dimension $k\ge 2$, then \begin{equation*} \limfunc{DIM}\nolimits_{\mathcal{C}}X=\limfunc{DIM}\nolimits_{\mathcal{D}}X=% \limfunc{DIM}\nolimits_{\mathrm{\limfunc{Ext}}}X=0